Question

Cho vecto a = (1;2), b =(-3;0), c = (-1;3)

a) Chứng minh hai vecto a, b không cũng phương

b) Phân tích vecto c qua vecto a, b 

Answer 1

Ta có: \(\dfrac{-3}{1}\ne\dfrac{0}{2}\Rightarrow\overrightarrow{a}\) và \(\overrightarrow{b}\) ko cùng phương

b. Đặt \(\overrightarrow{c}=x.\overrightarrow{a}+y.\overrightarrow{b}\)

\(\Rightarrow\left(-1;3\right)=x.\left(1;2\right)+y.\left(-3;0\right)=\left(x-3y;2x\right)\)

\(\Rightarrow\left\{{}\begin{matrix}x-3y=-1\\2x=3\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{3}{2}\\y=\dfrac{5}{6}\end{matrix}\right.\)

\(\Rightarrow\overrightarrow{c}=\dfrac{3}{2}\overrightarrow{a}+\dfrac{5}{6}\overrightarrow{b}\)